Question

The following data was collected from a simple random sample of a population.

Answer 1

Answer:

C

Step-by-step explanation:

This problem is analogous to the extraction of 6 elements from a total of 10 elements. It's the same if they are marbles, chips, or in this case, people, as here we don't care about the order of the selection as we only are drawing a sample.

Thus, the problem implies solving the amount of possible combinations of 10 people if we take by 6. There is a formula for this and is:

10 C 6 = 10!(6!4!)

If we operate, knowing that for any number x, x!=x*(x-1)*(x-2)*...*1

10 C 6 = 10!(6!4!) = 10*9*8*7*6*...*1 / [(6*5*...*1) * (4*3*2*1)]

10 C 6 = 10*9*8*7*6! / [(6*5*...*1) * (4*3*2*1)]

We have a 6! multiplying and another dividing, so they get eliminated, and as 4*2=8 and 9=3*3

10 C 6 = 10*9*8*7*/ [(4*3*2*1)] = 10*3*3*8*7*/ [(8*3*1)]

We can eliminate the 8s and one of the 3s on the numerator with the one on the denominator:

10 C 6 = 10*3*7*/1 = 210/1= 210

So, option C